Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Statistics
0 votes

Find the mean deviation about the mean for the data $36,72,46,42,60,45,53,46,51,49$

$\begin{array}{1 1}(A)\;3\\(B)\;4\\(C)\;7\\(D)\;6 \end{array} $

Can you answer this question?

1 Answer

0 votes
  • First arrange the given observations in ascending order and then calculate the median (n) after that mean deviation about the mean can be calculated by the reaction $ =\large\frac{\sum |x -M|}{n}$ where $n$=number of terms
Arranging the data in ascending order.
Number of observations $=10(even)$
Median $M=\Large\frac {N/2 observation + (N/2+1)th observation}{2}$
$\qquad= \Large\frac{(\large\frac{10}{2}) th \;observation + (\Large\frac{10}{2} +1)th \;observation}{2}$
$\qquad= \large\frac{5\;th\; observation +6th \;observation }{2}$
$\qquad= \large\frac{46+49}{2}$
$\qquad= 47.5 $
Step 2:
$X_i =36 , \qquad |X_i -M |=|36-47.5|=11.5$
$X_i =42 , \qquad |X_i -M |=|42-47.5|=5.5$
$X_i =45 , \qquad |X_i -M |=|45-47.5|=2.5$
$X_i =46 , \qquad |X_i -M |=|46-47.5|=1.5$
$X_i =46 , \qquad |X_i -M |=|46-47.5|=1.5$
$X_i =49 , \qquad |X_i -M |=|49-47.5|=2.5$
$X_i =51 , \qquad |X_i -M |=|51-47.5|=3.5$
$X_i =53 , \qquad |X_i -M |=|53-47.5|=5.5$
$X_i =60 , \qquad |X_i -M |=|60-47.5|=12.5$
$X_i =72 , \qquad |X_i -M |=|72-47.5|=24.5$
$\sum |x_i-M|=70$
Mean deviation about median $\large\frac{\sum|x_i-M|}{n}$
Hence C is the answer.
answered Jun 25, 2014 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App